Question

Prove that for all x and y in Z, x +3y is a multiple of 7 iff 3x +2y is a multiple of 7. Might be helpful to calculate 2(3x +2y)+(x +3y) and/or 4(x +3y)+(3x +2y).

Answer 1

Proof :

First, it is important to have in mind that a number
`m \in \mathbb{Z}` is a multiple of
`n\in\mathbb{Z}` iff there exists
`k\in\mathbb{Z}` such that
`m = n \cdot k`.

Also, you have to prove a logical equivalence. To this end, it is possible to prove two logical implications.

Explanation:

1.) Let x, y be integers such that x + 3y is a multiple of 7. You have to prove that 3x +2y is a multiple of 7.

In effect, by hypothesis there exists k
`\in\mathbb{Z}` such that x + 3y = 7 k . So, you get

`\begin{equation*} 4(x+3y) + (3x + 2y) = 7x + 14y = 7 (x + 2y) \ \mbox{(direct computations and factoring)}\end{equation*}`.

Therefore, 4(x +3y) + (3x +2y) is a multiple of 7. Then,

`(3x + 2y) = 7 (x + 2y) - 4(x + 3y) = 7 (x+2y) - 4 \cdot 7 k = 7 (x + 2y -4k) \ \mbox{(factoring)}`.

Given that x,y,k are integers, then x + 2y - 4k is an integer and hence, 3x + 2y is a multiple of 7.

To finish, it remains to prove its reciprocal statement.

2.) Let x, y be integers such that 3x + 2y is a multiple of 7. You have to prove that x +3y is a multiple of 7. Reasoining as before , there exists q
`\in\mathbb{Z}` such that 3x + 2y = 7 \cdot q. Thus,

`$$ \begin{equation*} 2(3x+2y) + (x + 3y) = 7x + 7y = 7 (x + y) \ \mbox{direct computations and factoring} \\\end{equation*} $$` Thus,
`2(3x +2y) + (x +3y)` is a multiple of 7.

On the other hand, using the hypothesis
`$$ \begin{equation*}   (x + 3y) = 7 (x + y) - 2(3x + 2y) = 7 (x+y) - 2 \cdot 7 q = 7 (x + y -2q) \ \mbox{(factoring)} \end{equation*} $$` .

Finally, thanks that
`x,y,q` are integer numbers, then
`x + y - 2q` is a integer number and therefore,
`3x + 2y` is a multiple of 7.